Answer:
Explanation:
Given that:
mass of the plastic ball = 50 g = 50 × 10⁻³ kg
spring constant (k) = 2.22 × 10⁵ N/m
compression of spring (x) = 3.0 cm = 3 × 10⁻² m
a) Work done:
[tex]= \dfrac{1}{2}kx^2 \\ \\ = \dfrac{1}{2}\times 2.22 \times 10^{5}\times (3 \times 10^{-2})^2 \\ \\ = 99.9 \ J[/tex]
b) When the gun is fired horizontally;
In the spring, the potential energy is changed to kinetic energy.
i.e.
[tex]\implies \dfrac{1}{2}kx^2 = \dfrac{1}{2}mv^2 \\ \\ 2.22 \times 10^5 \times (3\times 10^{-2}) ^2 =50 \times 10^{-3} \times v^2\\ \\ \mathbf{v= 63.2 m/s}[/tex]
c) The max height the bullet will reach if the gun is being shot upward is:
[tex]H_{max}= \dfrac{v^2}{2g} \\ \\ H_{max}= \dfrac{63.2^2}{2\times 9.8}[/tex]
[tex]\mathbf{H_{max}= 203.9 \ m}[/tex]
